Model Question Paper 5

A pentagonal pyramid is a pyramid whose base is a pentagon.

Its faces are the pentagonal base and the 5 triangular lateral faces.

Number of faces, F = 1 + 5 = 6

Its vertices are the 5 vertices of the pentagonal base and the apex.

Number of vertices, V = 5 + 1 = 6

Its edges are the 5 edges of the pentagonal base and the 5 slant edges.

Number of edges, E = 5 + 5 = 10

Hence, option 1 is the correct option.

For the point (-10, 0), the abscissa is -10 and the ordinate is 0.

Since the ordinate is 0, the point lies on the x-axis.

Since the abscissa is negative, the point lies to the left of the origin.

Hence, option 1 is the correct option.

Given,

Area of the semicircular region = 308 cm²

Let the radius of the semicircular region be $r$ cm.

By formula,

Area of a semicircle = $\frac{1}{2}πr^2$

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \frac{22}{7} \times r^2 = 308\\[1em] \Rightarrow \frac{11}{7} \times r^2 = 308\\[1em] \Rightarrow r^2 = \frac{308 \times 7}{11}\\[1em] \Rightarrow r^2 = 196\\[1em] \Rightarrow r = \sqrt{196}\\[1em] \Rightarrow r = 14\end{aligned}$$

So, radius of the semicircular region = 14 cm

Diameter of the semicircular region = 2 × radius

= 2 × 14 cm

= 28 cm

Hence, the diameter of the semicircular region is 28 cm.

Given,

Area of the trapezium = 96 cm²

Lengths of the two parallel sides = 14 cm and 10 cm

Let the perpendicular distance between the parallel sides be $h$ cm.

By formula,

Area of a trapezium = $\frac{1}{2}$ × sum of parallel sides × distance between the parallel sides

⇒ $\frac{1}{2} \times (14 + 10) \times h = 96$

⇒ $\frac{1}{2} \times 24 \times h = 96$

⇒ 12 × $h$ = 96

⇒ $h = \frac{96}{12}$

⇒ $h$ = 8 cm

Hence, the perpendicular distance between the parallel sides is 8 cm.

Given,

Radius of the circle = 6 cm

Length of the tangent from P to the circle = 8 cm

Let the tangent touch the circle at the point T.

As we know, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

So, ∠CTP = 90°.

In right-angled triangle CTP, CP is the hypotenuse.

By Pythagoras theorem,

⇒ CT² + TP² = CP²

⇒ (6)² + (8)² = CP²

⇒ 36 + 64 = CP²

⇒ CP² = 100

⇒ CP = $\sqrt{100}$

⇒ CP = 10 cm

The farthest point of the circumference from P lies on the line CP produced beyond the centre C, at a distance equal to the radius from C.

Distance of P from the farthest point of the circumference = CP + radius

= 10 + 6 cm

= 16 cm

Hence, the distance of the point P from the farthest point of the circumference is 16 cm.

Steps of construction:

1. Draw PQ = 5.5 cm.

2. Taking P as centre, draw an arc of radius 4.4 cm and taking Q as centre, draw one more arc of radius 7.1 cm. Let these two arcs intersect at point S.

3. Join PS and QS.

4. Taking P as centre, draw an arc of radius 6.5 cm on the same side of PS as Q and taking S as centre, draw one more arc of radius 6.5 cm. Let these two arcs intersect at point R.

5. Join PR, QR and SR.

Hence, PQRS is the required quadrilateral.

Given,

Length of the cuboidal water tank = 7 m

Breadth of the cuboidal water tank = 6 m

Height of the cuboidal water tank = 2 m

Volume of water pumped out = 4200 litres

Converting the volume of water into cubic metres,

Volume of water pumped out = $\frac{4200}{1000}$ m³

[∵ 1 m³ = 1000 litres]

= 4.2 m³

By formula,

Volume of a cuboid = length × breadth × height

Let the fall in the water level be h metres.

⇒ 7 × 6 × h = 4.2

⇒ 42 × h = 4.2

⇒ h = $\frac{4.2}{42}$

⇒ h = 0.1 m

Converting the fall in the water level into centimetres,

Fall in the water level = 0.1 × 100 cm = 10 cm

Hence, the fall in the water level in the tank is 10 cm.

We know that,

All the sides of a rhombus are equal.

So, PQ = QR = RS = SP = 5.4 cm

Steps of construction:

1. Draw PQ = 5.4 cm.

2. At P, construct ∠XPQ = 60°. From PX cut off PS = 5.4 cm.

3. Taking Q as centre, draw an arc of radius 5.4 cm (= QR) and taking S as centre, draw one more arc of radius 5.4 cm (= SR). Let these two arcs intersect at point R.

4. Join QR and SR.

A rhombus has two lines of symmetry — its diagonals PR and QS. They are shown by the dotted line in the figure above.

Hence, PQRS is the required rhombus and it has 2 lines of symmetry.

Given,

Diameter of the road roller = 84 cm

Width of the road roller = 1.2 m

Number of complete revolutions = 500

Rate of levelling = ₹2.50 per m²

Radius of the roller = $\frac{84}{2}$ cm = 42 cm = 0.42 m

In one complete revolution, the roller levels an area equal to its curved surface area.

As we know, curved surface area of a cylinder = 2πrh

$= 2 \times \frac{22}{7} \times 0.42 \times 1.2$

= 2 x 22 x 0.06 x 1.2

= 3.168

So, area levelled in one revolution = 3.168 m²

Area of the playground = Area levelled in one revolution × Number of revolutions

= 3.168 × 500 m²

= 1584 m²

Cost of levelling the playground = Area of the playground × Rate

= ₹1584 × 2.50

= ₹3,960

Hence, the area of the playground is 1584 m² and the cost of levelling it is ₹3,960.

We know that,

Each angle of a rectangle is a right angle and its opposite sides are equal.

So, ∠DAB = 90° and DC = AB = 6.4 cm.

Steps of construction:

1. Draw a line segment AB = 6.4 cm.

2. At A, construct ∠XAB = 90°.

3. Taking B as centre, draw an arc of radius 7.6 cm (= BD). Let it cut AX at point D.

4. At B, construct ∠YBA = 90°.

5. Taking D as centre, draw an arc of radius 6.4 cm (= DC). Let it cut BY at point C.

6. Join DC.

Hence, ABCD is the required rectangle.